Answer to Question #195617 in Calculus for Kobz

Question

A small funnel in the shape of a cone is being emptied of fluid at the rate of 12 cubic

centimeters per second. The height of the funnel is 20 centimeters, and the radius of the top is 4 centimeters. How fast is the fluid level dropping when the level stands 5 centimeters above the vertex of the cone?

Answer 1

Ans:-

We are given information about the rate of change of volume of water, so we are given that $\dfrac{dV} {dt} = −12$ . Note that the formula for volume is given in terms of r and h, but we only want $\dfrac{dh }{dt}$ . We need a relationship between r and h... You should see similar triangles (Draw a line right through the center of the cone. This, and the line forming the top radius are the two legs. The outer edge of the cone forms the hypotenuse).

$\dfrac{radius of top }{radius of water level} = \dfrac{overall height} {height of water} ⇒ \dfrac{4} {r} = \dfrac{20} {h}$

so that $r = \dfrac{h }{5}$ . Substituting this into the formula for the volume will give the volume in terms of h alone:

$V = \dfrac{1} 3 πr^2h = \dfrac{1} 3 π( \dfrac{h} 5 )^ 2h = \dfrac{1} {75} πh^3$

Now,

$\Rightarrow \dfrac{dV }{dt} = \dfrac{3π }{75 }h ^2 \dfrac{dh} {dt}$

and we know $\dfrac{dV} {dt} = −12, h = 5,$ So

$\Rightarrow \dfrac{dh}{ dt }= \dfrac{−12} π$